Question: You have found the following ages (in years) of all 6 meerkats at your local zoo: $ 7,\enspace 2,\enspace 1,\enspace 10,\enspace 10,\enspace 3$ What is the average age of the meerkats at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 meerkats at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{7 + 2 + 1 + 10 + 10 + 3}{{6}} = {5.5\text{ years old}} $ Find the squared deviations from the mean for each meerkat. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $1.5$ years $2.25$ years $^2$ $2$ years $-3.5$ years $12.25$ years $^2$ $1$ year $-4.5$ years $20.25$ years $^2$ $10$ years $4.5$ years $20.25$ years $^2$ $10$ years $4.5$ years $20.25$ years $^2$ $3$ years $-2.5$ years $6.25$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{2.25} + {12.25} + {20.25} + {20.25} + {20.25} + {6.25}} {{6}} $ $ {\sigma^2} = \dfrac{{81.5}}{{6}} = {13.58\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{13.58\text{ years}^2}} = {3.7\text{ years}} $ The average meerkat at the zoo is 5.5 years old. There is a standard deviation of 3.7 years.